Question

put this in a linear normal form (wait a bit for ques)

Answer 1

\[\sin(t)y'-\frac{ \log(t^2)+\frac{ y }{ t } }{ 1+e^{2t}}=y'\]

Answer 2

put that in the form of \[\frac{ dy }{ dt }+a(t)y=F(t)\]

Answer 3

factor out y' first

Answer 4

\[\large \sin(t)y'-\frac{ \log(t^2)+\frac{ y }{ t } }{ 1+e^{2t}}=y'\]

\[\large -\frac{ \log(t^2)+\frac{ y }{ t } }{ 1+e^{2t}}=y'(1-\sin t)\]

Answer 5

next isolate y term

Answer 6

\[\large \sin(t)y'-\frac{ \log(t^2)+\frac{ y }{ t } }{ 1+e^{2t}}=y'\]

\[\large -\frac{ \log(t^2)+\frac{ y }{ t } }{ 1+e^{2t}}=y'(1-\sin t)\]

\[\large -\frac{ \log(t^2)}{ 1+e^{2t}} -\frac{ \frac{ y }{ t } }{ 1+e^{2t}}=y'(1-\sin t)\]

Answer 7

divide through by (1-sint)

Answer 8

\[\large \sin(t)y'-\frac{ \log(t^2)+\frac{ y }{ t } }{ 1+e^{2t}}=y'\]

\[\large -\frac{ \log(t^2)+\frac{ y }{ t } }{ 1+e^{2t}}=y'(1-\sin t)\]

\[\large -\frac{ \log(t^2)}{ (1+e^{2t})((1-\sin t))} -\frac{ \frac{ y }{ t } }{ (1+e^{2t})((1-\sin t))}=y'\]

Answer 9

got this thanks

Answer 10

\[\large -\frac{ \log(t^2)}{ (1+e^{2t})((1-\sin t))} =y' + \frac{ y }{t (1+e^{2t})((1-\sin t))}\]